9.2 Calculating Acceleration (c) McGraw Hill Ryerson 2007

9.2 Calculating Acceleration The acceleration of an object depends on the change in velocity and the time required to change the velocity. When stopping a moving object, the relationship between time and acceleration is: Increasing the stopping time decreases the acceleration. Decreasing the stopping time increases the acceleration. Airbags cause the person to slow down in a longer period of time compared to hitting a solid object, such as the dashboard. This increased time results in a smaller deceleration. See page 392 (c) McGraw Hill Ryerson 2007

The slope of a velocity-time graph is average acceleration. Velocity-Time Graphs The motion of an object with uniform motion can be represented by a position-time graph. The motion of an object with a changing velocity can be represented by a velocity-time graph. The slope of a velocity-time graph is average acceleration. Acceleration is measured in m/s2. The slope of a velocity-time graph is the average acceleration of the object. See pages 393 - 394 (c) McGraw Hill Ryerson 2007

Determining Motion from a Velocity-Time Graph A velocity-time graph can be analyzed to describe the motion of an object. Positive slope (positive acceleration) – object’s velocity is increasing in the positive direction or the object’s velocity is decreasing in the negative direction Zero slope (zero acceleration) – object’s velocity is constant Negative slope (negative acceleration) – object’s velocity is increasing in the negative direction or the object’s velocity is decreasing in the positive direction State during which time interval: the acceleration was zero. the acceleration was negative. the acceleration was positive. the object was increasing its velocity north. the object was decreasing its velocity north. the object was moving at a constant velocity north. See pages 394 - 395 Answers are on the next slide (c) McGraw Hill Ryerson 2007

Determining Motion from a Velocity-Time Graph State during which time interval: the acceleration was zero. (t1 to t2) the acceleration was negative. (t2 to t3) the acceleration was positive. (0 to t1) the object was increasing it’s velocity north. (0 to t1) the object was decreasing it’s velocity north. (t2 to t3) the object was moving at a constant velocity north. (t1 to t2) See pages 394 - 395 (c) McGraw Hill Ryerson 2007

A velocity-time graph can be analyzed to describe the motion of an object. Positive slope (positive acceleration) – object’s velocity is increasing in the positive direction or the object’s velocity is decreasing in the negative direction Zero slope (zero acceleration) – object’s velocity is constant Negative slope (negative acceleration) – object’s velocity is increasing in the negative direction or the object’s velocity is decreasing in the positive direction State during which time interval: the acceleration was zero. the acceleration was negative. the acceleration was positive. the object was increasing its velocity south. the object was decreasing its velocity south. the object was moving at a constant velocity south. See pages 394 - 395 Answers are on the next slide (c) McGraw Hill Ryerson 2007

Determining Motion from a Velocity-Time Graph State during which time interval: the acceleration was zero. (t1 to t2) the acceleration was negative. (0 to t1) the acceleration was positive. (t2 to t3) the object was increasing it’s velocity south. (0 to t1) the object was decreasing it’s velocity south. (t2 to t3) the object was moving at a constant velocity south. (t1 to t2) See pages 394 - 395 (c) McGraw Hill Ryerson 2007

Calculating Acceleration The relationship of acceleration, change in velocity, and time interval is given by the equation: Example: A pool ball travelling at 2.5 m/s towards the cushion bounces off at 1.5 m/s. If the ball was in contact with the cushion for 0.20 s, what is the ball’s acceleration? (Assume towards the cushion is the positive direction. (+)) The ball’s velocity changes from 2.5 m/s toward the cushion (A) to 1.5 m/s away from the cushion (B). See pages 396 - 397 (c) McGraw Hill Ryerson 2007

Calculating Acceleration The relationship of change in velocity, acceleration, and time interval is given by the equation: Example: A car accelerates from rest at 3.0 m/s2 forward for 5.0 s. What is the velocity of the car at the end of 5.0 s? The car accelerates from rest for 5.0 s. The car’s change in velocity is 15 m/s forward, therefore See pages 396 - 397 The car’s velocity after 5.0 s is 15 m/s forward. (c) McGraw Hill Ryerson 2007

Calculating Acceleration The relationship of time interval, change in velocity, and acceleration is given by the equation: Example: A train is travelling east at 14 m/s. How long would it take to increase its velocity to 22 m/s east, if it accelerated at 0.50 m/s2 east? Assign east direction positive (+). To find the value of Δt: See pages 396 - 397 It would take 16 s for the train to increase its velocity. (c) McGraw Hill Ryerson 2007

Calculating Acceleration Try the following acceleration problems. A truck starting from rest accelerates uniformly to 18 m/s [W] in 4.5 s. What is the truck’s acceleration? How much time does it take a car travelling south at 12 m/s to increase its velocity to 26 m/s south if it accelerates at 3.5 m/s2 south? A toboggan moving 5.0 m/s forward decelerates backward at -0.40 m/s2 for 10 s. What is the toboggan’s velocity at the end of the 10 s? Answers are on the next slide. See page 397 (c) McGraw Hill Ryerson 2007

Calculating Acceleration (continued) Try the following acceleration problems. A truck starting from rest accelerates uniformly to 18 m/s [W] in 4.5 s. What is the truck’s acceleration? (4.0 m/s2 [W]) How much time does it take a car travelling south at 12 m/s to increase its velocity to 26 m/s south if it accelerates at 3.5 m/s2 south? (4.0 s) A toboggan moving 5.0 m/s forward decelerates backward at -0.40 m/s2 for 10 s. What is the toboggan’s velocity at the end of the 10 s? (1.0 m/s forward) See page 397 (c) McGraw Hill Ryerson 2007

Complete pages 172 – 173 of your work book. Work Time! With a partner, do activity 9-2B on page 396 of your textbook. Compare your answers to another group. Does everyone agree on the correct answers? Complete pages 172 – 173 of your work book. See pages 398 - 399 (c) McGraw Hill Ryerson 2007

9.2 Gravity and Acceleration (c) McGraw Hill Ryerson 2007

Gravity and Acceleration Objects near the surface of Earth fall to Earth due to the force of gravity. Gravity is a pulling force that acts between two or more masses. Air resistance is a friction-like force that opposes the motion of objects that move through the air. Ignoring air resistance, all objects will accelerate towards Earth at the same rate. The acceleration due to gravity is 9.8 m/s2 downward. See pages 398 - 399 (c) McGraw Hill Ryerson 2007

Calculating Motion Due to Gravity To analyze situation where objects are accelerating due to gravity, use the equations: In these equations, the acceleration ( ) is 9.8 m/s2 downward. Example: Suppose a rock falls from the top of a cliff. What is the change in velocity of the rock after it has fallen for 1.5 s? Assign “down” as negative (-). Since down is negative (-), the change in the rock’s velocity is 15 m/s down. See page 400 (c) McGraw Hill Ryerson 2007

Calculating Motion Due to Gravity Try the following acceleration due to gravity problems. What is the change in velocity of a brick that falls for 3.5 s? A ball is thrown straight up into the air at 14 m/s. How long does it take for the ball to slow down to an upward velocity of 6.0 m/s? A rock is thrown downwards with an initial velocity of 8.0 m/s. What is the velocity of the rock after 1.5 s? See page 400 Answers are on the next slide. (c) McGraw Hill Ryerson 2007

Calculating Motion Due to Gravity (continued) Try the following acceleration due to gravity problems. What is the change in velocity of a brick that falls for 3.5 s? (34.3 m/s downward) A ball is thrown straight up into the air at 14 m/s. How long does it take for the ball to slow down to an upward velocity of 6.0 m/s? (0.8 s) A rock is thrown downwards with an initial velocity of 8.0 m/s. What is the velocity of the rock after 1.5 s? (22.7m/s downward) See page 400 (c) McGraw Hill Ryerson 2007

Do questions # 1 – 13 on page 405 of your textbook Work Time! Do questions # 1 – 13 on page 405 of your textbook Complete pages 174 – 176 of your work book. Challenge Question: Explain in words what is happening to the motion of the object in this velocity-time graph. See pages 398 - 399 (c) McGraw Hill Ryerson 2007